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2 years ago

Explain why solving (2/5)c = (8/9) by multiplying both sides by (5/2) is the same as solving it by dividing both sides by (2/5)

1 answer:

3 0

Dividing by a fraction is the same as multiplying by its reciprocal. So dividing by 2/5 will give you the same answer as multiplying by 5/2.

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A cone has a radius of 8 inches and a height of 12 inches.What is the exact volume of the cone?

Cerrena [4.2K]

$The\ volume\ of\ the\ cone:V=\frac{1}{3}\pi r^2H\\\\where\\r-radius\\H-height\\----------------\\\\r=8\ in\ and\ H=12\ in\\\\therefore\\\\V=\frac{1}3{\pi\cdot8^2\cdot12=\frac{1}{\not3_1}\cdot\frac{\not12^4}{1}\cdot64\pi=4\cdot64\pi=\boxed{256\pi\ (in^3)}$

8 0

2 years ago

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Plz help me asap i need help plz i will mark you as brainlist

geniusboy [140]

Answer:

the third one

6 0

2 years ago

The circumference of a circle is 15π centimeters. What is the area of the circle in terms of π?

miss Akunina [59]

Answer: 7.5²π square centimeters. OR 56.25π cm²

Step-by-step explanation:

Circumference = diameter × π

So the diameter is 15 cm

Area = πr²

Radius is half the diameter, so 15/2 or 7.5

Area = (15/2)²π cm²

7 0

2 years ago

Find dy/dx for y= x^3 ln (cot x)

ICE Princess25 [194]

<h3>Answer</h3>

$\dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)$

<h3>Explanation</h3>

By the product rule $(d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$ , we have

$\begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}$

By the chain rule:

$\begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}$

By the power rule:

$(x^3)' = 3x^2$

thus

$\begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}$

Nothing to do to simplify any further, other than factoring out $x^2$ .

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2 years ago

Anyone waanna talk on Instagram

Maslowich

Ok sure and one more thing are you smart at geometry like 100% smart

8 0

2 years ago

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